Question

Consider a maximization problem with the optimal tableau in Table 73. The optimal solution to this LP is z = 10, x3 = 3, x4 = 5, x1 = x2 = 0. Determine the second-best bfs to this LP. (Hint: Show that the second-best solution must be a bfs that is one pivot away from the optimal solution.) TABLE 73 z X1 X2 X3 X4 rhs 1 2 10 10 10 0 3 2 1 0 3 0 430 11

please Solve This!!

Answer 1

Sice optimal table is

Z	x1	x2	x3	x4	rhs
1	2	1	0	0	10
0	3	2	1	0	3
0	4	3	0	1	5

Optimal solution is z=10 for x1=x2=0, x3=3,x4=5

To find the second best basic feasible solution i.e.bfs

One pivot away from the optimal solution we choose x2 as the entering variable bcoz it gives the smallest decrease in objective value z and corresponding leaving variable is x3 according to the ratio test

After two iteration new table is determined by applying

$R2\rightarrow 1/2R2$

R1 + R1(old) – R2(new)

$R3\rightarrow R3-3R2(new)$

z	x1	x2	x3	x4	rhs
1	1/2	0	-1/2	0	17/2
0	3/2	1	1/2	0	3/2
0	-1/2	0	-3/2	1	1/2

So second best bfs is x2=3/2 ; x4=1/2 ;z=17/2